HSC 2013 MX2 Marathon (archive) (2 Viewers)

seanieg89 · Nov 14, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Suppose three complex numbers $z_1~z_2~z_3$ make an equilateral triangle on an argand diagram. Show that:

$z_1^2 + z_2^2 + z_3^2 = |z_1||z_2| + |z_1||z_3| + |z_2||z_3|$

remove modulus signs.

RealiseNothing · Nov 14, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
remove modulus signs.

Thanks for that, I couldn't remember the exact condition/equation.

seanieg89 · Nov 14, 2012

Re: HSC 2013 4U Marathon

Also, more interestingly, the condition is equivalent to the three points forming an equilateral triangle. (If and only if).

cutemouse · Nov 14, 2012

Re: HSC 2013 4U Marathon

I think an 'elegant' approach to this is setting one of the points to 0, finding a condition and then using a change of coordinates.

Carrotsticks · Nov 14, 2012

Re: HSC 2013 4U Marathon

cutemouse said:
I think an 'elegant' approach to this is setting one of the points to 0, finding a condition and then using a change of coordinates.

That is the fastest method that I know of at the moment, though there are most likely faster ones. A typical 'school test proof' would be to actually draw out arbitrary vectors z_1, z_2 and z_3 then work with the condition, though this causes problems because it only proves one direction of the statement.

w.l.o.g you can shift one of the points z_1, z_2 or z_3 to the origin, then solve the quadratic. Upon doing so, you should immediately acquire that the other two are plus/minus cis(pi/3) away whilst preserving modulus. Since the solving of the quadratic works both ways, you have consequently proved both directions of the 'iff' statement.

seanieg89 · Nov 14, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
That is the fastest method that I know of at the moment, though there are most likely faster ones. A typical 'school test proof' would be to actually draw out arbitrary vectors z_1, z_2 and z_3 then work with the condition, though this causes problems because it only proves one direction of the statement.

w.l.o.g you can shift one of the points z_1, z_2 or z_3 to the origin, then solve the quadratic. Upon doing so, you should immediately acquire that the other two are plus/minus cis(pi/3) away whilst preserving modulus. Since the solving of the quadratic works both ways, you have consequently proved both directions of the 'iff' statement.

That used to be my fastest method. I just thought of something arguably faster though.

Let a=z1-z2, b=z2-z3, c=z3-z1. Observe that a+b+c=0 always.

The equation given is equivalent to a^2+b^2+c^2=0. But since a+b+c is also zero, this is equivalent to the fact that the monic polynomial with roots a,b,c must have zero coefficients for z and z^2. This is equivalent to a,b and c having the same modulus. This is equivalent to the triangle with vertices z1,z2,z3 being equilateral.

seanieg89 · Nov 14, 2012

Re: HSC 2013 4U Marathon

They are roughly equal in length probably.

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

Well that was an excellent question.

Solution

I feel like my solution is unnecessarily long however.

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Given: $ \sum_{i=1}^j i^2 = \frac{j}{6}(j+1)(2j+1)$

$Show that $ \sum_{r=1}^n (\left \sum_{k=1}^r k)\right = \frac{n}{6}(n+1)(n+2) \\ \\ $Hence prove that the product of any three consecutive integers is divisible by 6$

Note:

$\sum_{k=1}^r k = \frac{r}{2}(r+1)$

(Arithmetic series)

$\sum_{r=1}^n \frac{r}{2}(r+1) \\ \\ = 0.5 \sum_{r=1}^n r^2 +r \\ \\ = 0.5 [\sum_{r=1}^n r^2 + \sum_{r=1}^n r] \\ \\ = 0.5[\frac{n}{6}(n+1)(2n+1)+\frac{n}{2}(n+1)] \\ \\ = 0.5(n/6 (n+1) (2n+1+3) = \frac{n}{6}(n+1)(n+2)$

EDIT: If you want to read about this
http://en.wikipedia.org/wiki/Figurate_number

==============

$$Given the polynomial $ x^3-3x^2+5x+2012 $ has roots $ \alpha, \beta, \gamma$

$$i) Find $ (\alpha-1)(\beta-1)(\gamma -1)$

$$ii) Hence or otherwise find $ (\beta+ \gamma - \alpha)(\alpha+\beta-\gamma)(\alpha+\gamma -\beta)$

twinklegal19 · Nov 14, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Note:

$\sum_{k=1}^r k = \frac{r}{2}(r+1)$

(Arithmetic series)

$\sum_{r=1}^n \frac{r}{2}(r+1) \\ \\ = 0.5 \sum_{r=1}^n r^2 +r \\ \\ = 0.5 [\sum_{r=1}^n r^2 + \sum_{r=1}^n r] \\ \\ = 0.5[\frac{n}{6}(n+1)(2n+1)+\frac{n}{2}(n+1)] \\ \\ = 0.5(n/2 (n+1) (\frac{2n+4}{3}) = \frac{n}{6}(n+1)(n+2)$

==============

$$Given the polynomial $ x^3-3x^2+5x+2012 $ has roots $ \alpha, \beta, \gamma$

$$i) Find $ (\alpha-1)(\beta-1)(\gamma -1)$

$$ii) Hence or otherwise find $ (\beta+ \gamma - \alpha)(\alpha+\beta-\gamma)(\alpha+\gamma -\beta)$

$\\x^3-3x^2+5x+2012=0\Rightarrow x=\alpha,\beta,\gamma\\\sum \alpha=3\\\sum \alpha\beta=-5\\\sum\alpha\beta\gamma=-2012\\$i. $(\alpha-1)(\beta-1)(\gamma -1)\\=\alpha\beta\gamma-(\alpha\beta+\beta\gamma+\alpha\gamma)+(\alpha+\beta+\gamma)-1\\=(-2012)-(-5)+(3)-1)\\=-2005\\$ii. $(\beta+ \gamma - \alpha)(\alpha+\beta-\gamma)(\alpha+\gamma -\beta)\\=(3-2\alpha)(3-2\beta)(3-2\gamma)\\=-8(\alpha\beta\gamma)+12(\alpha\beta+\beta\gamma+\alpha\gamma)-18(\alpha+\beta+\gamma)+27\\=-8(-2012)+12(-5)-18(3)+27\\=16009$

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
$\\x^3-3x^2+5x+2012=0\Rightarrow x=\alpha,\beta,\gamma\\\sum \alpha=3\\\sum \alpha\beta=-5\\\sum\alpha\beta\gamma=-2012\\$i. $(\alpha-1)(\beta-1)(\gamma -1)\\=\alpha\beta\gamma-(\alpha\beta+\beta\gamma+\alpha\gamma)+(\alpha+\beta+\gamma)-1\\=(-2012)-(-5)+(3)-1)\\=-2005\\$ii. $(\beta+ \gamma - \alpha)(\alpha+\beta-\gamma)(\alpha+\gamma -\beta)\\=(3-2\alpha)(3-2\beta)(3-2\gamma)\\=-8(\alpha\beta\gamma)+12(\alpha\beta+\beta\gamma+\alpha\gamma)-18(\alpha+\beta+\gamma)+27\\=-8(-2012)+12(-5)-18(3)+27\\=16009$

Ah my second question was not built nicely. Sorry about that. (Was supposed to lead on from part i)
It should of been.
$x^3-2x^2+5x+2012$

Will think of another question soon.

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

You are given:

$0<a_0, a_1, a_2, a_3 .... ,a_n < 10 \ \ \ \ \ \ \ \ \ a_0, a_1, a_2, .... , a_n \in \mathbb{N}$

$a_0+a_1+a_2+...+a_n=3p \ \ \ \ \ p \in \mathbb{N}$

$Prove by process of mathematical induction or otherwise$

$10^n a_n+10^{n-1}a_{n-1}+10^{n-2}a_{n-2}+...+1000a_3+100a_2+10a_1+a_0 = 3k \ \ \ \ k \in \mathbb{N}$

(So basically prove that expression is divisible by 3)

RealiseNothing · Nov 14, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
You are given:

$0<a_1, a_2, a_3 .... ,a_n < 10 \ \ \ \ \ \ \ \ \ a_1, a_2, .... , a_n \in \mathbb{N}$

$a_1+a_2+...+a_n=3p \ \ \ \ \ p \in \mathbb{N}$

$Prove by process of mathematical induction or otherwise$

$10^n a_n+10^{n-1}a_{n-1}+10^{n-2}a_{n-2}+...+100a_3+10a_2+a_1 = 3k \ \ \ \ k \in \mathbb{N}$

(So basically prove that expression is divisible by 3)

You sure that's right?

The general term is $10^ka_k$ so shouldn't it be $1000a_3$ and $100a_2$ and $10a_1$ ?

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
You sure that's right?

The general term is $10^ka_k$ so shouldn't it be $1000a_3$ and $100a_2$ and $10a_1$ ?

Apologies, will fix it now

RealiseNothing · Nov 14, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
You are given:

$0<a_0, a_1, a_2, a_3 .... ,a_n < 10 \ \ \ \ \ \ \ \ \ a_0, a_1, a_2, .... , a_n \in \mathbb{N}$

$a_0+a_1+a_2+...+a_n=3p \ \ \ \ \ p \in \mathbb{N}$

$Prove by process of mathematical induction or otherwise$

$10^n a_n+10^{n-1}a_{n-1}+10^{n-2}a_{n-2}+...+1000a_3+100a_2+10a_1+a_0 = 3k \ \ \ \ k \in \mathbb{N}$

(So basically prove that expression is divisible by 3)

I'm going to skip the "prove true for n=0" bits as they are just trivial really.

Well we can assume:

$10^na_n + ... + a_0=3k$

Now we want to prove that:

$10^{n+1}a_{n+1} + 10^na_n + ... + a_0 = 3m$

By substituting in our assumption:

$10^{n+1}a_{n+1} + 3k$

Now we are given that:

$a_0 + a_1 +...+ a_n = 3p$

Hence:

$a_{n+1} = 3q - a_n - a_{n-1} - ... - a_1 - a_0$

Substituting this in gives us:

$10^{n+1}(3q - a_n - a_{n-1} - ... - a_1 - a_0) + 3k$

$10^{n+1}(3q - (a_n + a_{n-1} + ... + a_1 + a_0)) + 3k$

But we were given:

$a_n + a_{n-1} + ... + a_1 + a_0 = 3p$

By substituting this in:

$10^{n+1}(3q - 3p) + 3k$

$3(q - p)10^{n+1} + 3k$

$3((q - p)10^{n+1} + k)$

$=3m$ where $m=((q - p)10^{n+1} + k)$

Therefore it is divisible by 3 for all blah blah, etc.

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
I'm going to skip the "prove true for n=0" bits as they are just trivial really.

Well we can assume:

$10^na_n + ... + a_0=3k$

Now we want to prove that:

$10^{n+1}a_{n+1} + 10^na_n + ... + a_0 = 3m$

By substituting in our assumption:

$10^{n+1}a_{n+1} + 3k$

Now we are given that:

$a_0 + a_1 +...+ a_n = 3p$

Hence:

$a_{n+1} = 3q - a_n - a_{n-1} - ... - a_1 - a_0$

Substituting this in gives us:

$10^{n+1}(3q - a_n - a_{n-1} - ... - a_1 - a_0) + 3k$

$10^{n+1}(3q - (a_n + a_{n-1} + ... + a_1 + a_0)) + 3k$

But we were given:

$a_n + a_{n-1} + ... + a_1 + a_0 = 3p$

By substituting this in:

$10^{n+1}(3q - 3p) + 3k$

$3(q - p)10^{n+1} + 3k$

$3((q - p)10^{n+1} + k)$

$=3m$ where $m=((q - p)10^{n+1} + k)$

Therefore it is divisible by 3 for all blah blah, etc.

Nice work. I will post another question if I can think of one (unless someone else can post one?)

RealiseNothing · Nov 14, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Nice work. I will post another question if I can think of one (unless someone else can post one?)

I would post one, but I'm saving them for the little complex numbers quiz I'm writing lol.

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
I would post one, but I'm saving them for the little complex numbers quiz I'm writing lol.

Alright haha. Ill post another one tomorrow (I will try to make it a Complex Numbers/Polynomials one)

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

$$Given the polynomial $ S(x)=ax^3+bx^2 + bx+a$

$The polynomial has 3 real roots all of which have a multiplicity of 2 or less.$

$i) Factorise the polynomial as one linear and one quadratic factor$

$$ii) Show that if the polynomial has 3 real roots then $ b \geq 3a$

$$iii) What happens when $ b=3a$

twinklegal19 · Nov 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Given the polynomial $ S(x)=ax^3+bx^2 + bx+a$

$The polynomial has 3 real roots all of which have a multiplicity of 2 or less.$

$i) Factorise the polynomial as one linear and one quadratic factor$

$$ii) Show that if the polynomial has 3 real roots then $ b \geq 3a$

$$iii) What happens when $ b=3a$

~~still editing~~

$\\$i) by inspection, $(x+1)$ is a factor$\\S(x)=(x+1)(ax^2+cx+a)\\$equating coefficients of $x\\b=a+c\\c=b-a\\S(x)=(x+1)[ax^2+(b-a)x+a]$
$\\ax^2+(b-a)x+a=0\\\bigtriangleup =(b-a)^2-4a^2\\$For real roots, \bigtriangleup \geq 0$\\(b-a)^2-4a^2\geq 0\\(b-a-2a)(b-a+2a)\geq 0\\(b-3a)(b+a)\geq 0$

Edit: sorry got to study for an English exam omorrow, someone feel free to take over lol.

HSC 2013 MX2 Marathon (archive) (2 Viewers)

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